Java Program For Selecting A Random Node From A Singly Linked List

Given a singly linked list, select a random node from the linked list (the probability of picking a node should be 1/N if there are N nodes in the list). You are given a random number generator.
Below is a Simple Solution:

1. Count the number of nodes by traversing the list.
2. Traverse the list again and select every node with probability 1/N. The selection can be done by generating a random number from 0 to N-i for i’th node, and selecting the i’th node only if the generated number is equal to 0 (or any other fixed number from 0 to N-i).

We get uniform probabilities with the above schemes.  

i = 1, probability of selecting first node = 1/N
i = 2, probability of selecting second node =
                   [probability that first node is not selected] * 
                   [probability that second node is selected]
                  = ((N-1)/N)* 1/(N-1)
                  = 1/N  

Similarly, probabilities of other selecting other nodes is 1/N
The above solution requires two traversals of linked list. 

How to select a random node with only one traversal allowed?
The idea is to use Reservoir Sampling. Following are the steps. This is a simpler version of Reservoir Sampling as we need to select only one key instead of k keys.

(1) Initialize result as first node
    result = head->key 
(2) Initialize n = 2
(3) Now one by one consider all nodes from 2nd node onward.
    (a) Generate a random number from 0 to n-1. 
        Let the generated random number is j.
    (b) If j is equal to 0 (we could choose other fixed numbers 
        between 0 to n-1), then replace result with the current node.
    (c) n = n+1
    (d) current = current->next

Below is the implementation of above algorithm.

Java

// Java program to select a random
// node from singly linked list
import java.util.*;

// Linked List Class
class LinkedList
{
    // head of list
    static Node head;

    // Node Class
    static class Node
    {
        int data;
        Node next;

        // Constructor to create
        // a new node
        Node(int d)
       {
            data = d;
            next = null;
        }
    }

    // A reservoir sampling-based function
    // to print a random node from a
    // linked list
    void printrandom(Node node)
    {
        // If list is empty
        if (node == null)
        {
            return;
        }

        // Use a different seed value so
        // that we don't get same result
        // each time we run this program
        Math.abs(UUID.randomUUID().getMostSignificantBits());

        // Initialize result as first node
        int result = node.data;

        // Iterate from the (k+1)th element
        // to nth element
        Node current = node;
        int n;
        for (n = 2; current != null; n++)
        {
            // change result with
            // probability 1/n
            if (Math.random() % n == 0)
            {
                result = current.data;
            }

            // Move to next node
            current = current.next;
        }
        System.out.println(
        "Randomly selected key is " + result);
    }

    // Driver code
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
        list.head = new Node(5);
        list.head.next = new Node(20);
        list.head.next.next = new Node(4);
        list.head.next.next.next = new Node(3);
        list.head.next.next.next.next = new Node(30);
        list.printrandom(head);
    }
}
// This code is contributed by Mayank Jaiswal

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.

Note that the above program is based on the outcome of a random function and may produce different output.

How does this work?
Let there be total N nodes in list. It is easier to understand from the last node.
The probability that the last node is result simply 1/N [For last or N’th node, we generate a random number between 0 to N-1 and make the last node as a result if the generated number is 0 (or any other fixed number]
The probability that second last node is result should also be 1/N. 

The probability that the second last node is result 
          = [Probability that the second last node replaces result] X 
            [Probability that the last node doesn't replace the result] 
          = [1 / (N-1)] * [(N-1)/N]
          = 1/N

Similarly, we can show probability for 3^rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!